I Can we order Complex Numbers?

1. Apr 1, 2017

Arman777

Can we order Complex Numbers ? I searched a bit most places says it can but not like the real numbers. I am confused a bit.And I am not sure abouth the truth of those sources.
Thanks

2. Apr 1, 2017

zwierz

We can order every set. But the question is will such an order have any relation to a complex structure or even to linear or topological structure in $\mathbb{R}^2$ :)

3. Apr 1, 2017

JoePhysics

If I recall correctly, the way it works is that if you try impose an order in the complex numbers, you then lose the field properties.

4. Apr 1, 2017

Staff: Mentor

The complex numbers don't allow an Archimedean order. This is equivalent to the condition that squares are positive, which is not the case for complex numbers, as $i^2=-1<0$. They allow however an order like the lexicographical order: $x+iy < u+iv \Longleftrightarrow x < u \,\vee \, (x=u \wedge y < v)$.

5. Apr 1, 2017

Math_QED

If you mean ordering like we think of ordering with real numbers, the answer is no. To demonstrate this, consider this example that uses an informal argument.

If $\mathbb{C}$ is a totally ordered set with properties like the real numbers, we have the axiom of totality, which says that $\forall x,y \in \mathbb{C}: x \leq y$ or $x \geq y$

Thus, consider $i \leq 0$, then multiply both sides with $i$, obtaining $-1 \geq 0$ (multiplying with a number less than $0$ reverses the inequality sign). This is not possible.
Otherwise, assume $i \geq 0$, then $-1 \geq 0$. Also impossible.

Last edited: Apr 1, 2017
6. Apr 1, 2017

Arman777

I understand thanks

7. Apr 1, 2017

Staff: Mentor

You can find a bijective function between the complex numbers and the real numbers and then use the ordering of the real numbers to "order" the complex numbers. That won't lead to nice properties of the ordering, but it is some sort of order, similar to the one @fresh_42 suggested.

You can even have a well-ordering, which places stronger conditions on an order.